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Evaluation of Integral $\displaystyle \int \sqrt{\sin x}\; dx$

$\bf{My\; Try::}$ Let $\sin x = y^2\;, $ Then $\displaystyle \cos xdx =2ydy\Rightarrow dx = \frac{2y}{\sqrt{1-y^4}}dy$

So $\displaystyle \int \sqrt{\sin x}\;dx = 2\int \frac{y^2}{\sqrt{1-y^4}}dy = 2\int y^2\cdot \left(1-y^4\right)^{-\frac{1}{2}}dy$

(Using Wolframalpha It Show The Results is in the form of Elleptical Integral of first and Second Kind.)

Now How Can I solve after that

Help Required



marked as duplicate by Eric Towers, Claude Leibovici, Ted, mau, Cookie Jun 29 '14 at 8:51

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Effectively $$\displaystyle \begin{align} & \int \sqrt{\sin x}\; dx=-2 E\left(\left.\frac{1}{4} (\pi -2 x)\right|2\right) \\ & 2\int \frac{y^2}{\sqrt{1-y^4}}dy= 2 \left(E\left(\left.\sin ^{-1}(y)\right|-1\right)-F\left(\left.\sin ^{-1}(y)\right|-1\right)\right) \end{align}$$ In my opinion, you can not get rid of the elliptic integrals. The result would be still worse using Weierstrass substitution.

  • $\begingroup$ Mr. Claude, you can always try to align LaTeX equations by using \begin{align} & Equation \\ & II Equation \end{align} . This looks much better. :) $\endgroup$ – Kushashwa Ravi Shrimali Jun 29 '14 at 6:29
  • $\begingroup$ @KushashwaRaviShrimali. Thank you very much ! Being almost blind, editing with LaTeX is very hard for me even if I feel I improved a lot during the last months. Cheers :) $\endgroup$ – Claude Leibovici Jun 29 '14 at 6:48
  • $\begingroup$ Mr. Claude, you're most welcome. I understand and its okay! Thanks for appreciating this. By the way, you have been such a great asset on SE and helping others is much better than editing the work :) $\endgroup$ – Kushashwa Ravi Shrimali Jun 29 '14 at 6:55

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